AP Physics B & C Kinematics- More Multiple Choice Questions on One Dimensional Motion

The best thinking has been done in solitude. The worst has been done in turmoil

– Thomas Alva Edison

The essential things you need to remember in one dimensional uniformly accelerated motion were discussed in the post dated 13^th August 2008. A few multiple choice questions on one dimensional motion were discussed in the post dated 21^st August 2008. You can access both posts by clicking on the label ‘one dimensional motion’ below this post. Today we will discuss a few more multiple choice practice questions in this section:

(1) In the straight line graph AB shown in the adjoining figure time is taken along the x-axis. The quantity taken along the y-axis (which is not shown) in the graph can be

(a) acceleration of a train arriving at a railway station

(b) velocity of a train arriving at a railway station

(c) velocity of a cricket ball thrown towards a batsman

(d) acceleration of a ball thrown vertically downwards

(e) velocity of a ball thrown vertically upwards

When a ball is thrown vertically upwards, its velocity v goes on decreasing linearly with time t in accordance with the equation,

v = v₀ – gt where v₀ is the velocity of projection and g is the acceleration due to gravity.

At the maximum height the velocity is zero. Thereafter the direction of the velocity is reversed (during the return trip of the ball) and the magnitude of velocity goes on increasing linearly with time in accordance with the equation,

v = v₀ + gt

Therefore, the graph AB is the velocity-time graph of a body projected vertically upwards. Option (e) is correct.

(2) An iron sphere released from rest at a height h falls freely under gravity. It reaches half the height at the instant t₁ and hits the ground at the instant t₂. if air resistance is negligible, the ratio t₁/t₂ is

(a) 2^1/2

(b) 2^–1/2

(c) 2^–1/4

(d) 2^1/4

(e) ¼

We have s = v₀ t+ (½) at² where s is the displacement in time t, v₀ is the initial velocity and a is the acceleration.

Since the displacements in times t₁ and t₂ are h/2 and h respectively, we have

h/2 = 0 + (½) g t₁² and

h= 0 + (½) g t₂²

Therefore t₁²/t₂² = ½ so that t₁/t₂ = 2^–1/2

(3) A rifle can fire bullets with a speed of 1000 ms^–1. Assume that air resistance is negligible and the gravitational acceleration is 10 ms^–2. In order to hit a target at a horizontal distance of 100 m, the rifle is to be aimed

(a) 5 cm above the target

(b) 10 cm above the target

(c) 10 cm below the target

(d) 5 cm below the target

(e) directly at the target

Since the gravitational force does not affect the horizontal velocity of the bullet, the time t taken to cover the horizontal distance of the target is given by

t = 100/1000 = 0.1 s

During this time the bullet will fall down by a distance s given by

s = ½ gt² = ½ ×10×(0.1)² = 0.05 m = 5 cm.

Therefore, in order to hit the target, the rifle is to be aimed at a point 5 cm above the target [Option (a)].

(4) When a body is thrown vertically down with a velocity of 4 ms^–1 from a height h it reaches the ground with a velocity of 5 ms^–1. If the same body is is released from rest from the same height h, it it will reach the ground with a velocity

(a) 2 ms^–1

(b) 2.5 ms^–1

(c) 3 ms^–1

(d) 3.5 ms^–1

(e) 4 ms^–1

We can use here the equation of uniformly accelerated motion,

v² = u² + 2as where u is the initial velocity and v is the final velocity after a body moving with an acceleration a suffers a displacement s.

When the body is thrown down with a velocity of 4 ms^–1 from a height h we have

5² = 4² + 2gh …….(i)

where g is the acceleration due to gravity.

When the body is released from rest from the same height h we have

v² = 0 + 2gh ……..(ii)

Subtracting Eq.(ii) from Eq.(i) we obtain

5² – v² = 4² from which v = 3 ms^–1.

The following questions [(5) and (6)] are meant for AP Physics C aspirants:

(5) A particle moving along the positive x-direction has its position coordinate x at time t given by

√x = b t+ c where b and c are constants.

The acceleration and the initial velocity of the particle are respectively

(a) zero, b

(b) zero, 2bc

(c) b², bc

(d) b², 2bc

(e) 2b², 2bc

Squaring the equation for the position coordinate, we have

x = b²t² + c² + 2bct

The velocity v of the particle at the instant t is given by

v = dx/dt = 2 b²t + 2bc

Or, v =2bc + 2b²t

When t = 0, v = 2bc. This is the initial velocity of the particle.

The acceleration a of the particle is given by

a = dv/dt = 2b²

[You can obtain the answers from the form of the equation, x = b²t² + c² + 2bct itself. This equation is in the form, x = x₀ + v₀t + ½ at² which gives the displacement (x – x₀) at the instant t of a particle moving along a straight line with acceleration a and initial velocity v₀].

(6) A ball projected vertically up from the ground moves past a point P at a height h at the end of 2 seconds during the upward trip. After another 2 seconds it moves past the same point P during the downward trip. Neglect air resistance and assume that g = 10 ms^–2. The maximum height reached by the ball is

(a) 50 m

(b) 45 m

(c) 40 m

(d) 33 m

(e) 22 m

The time taken by the ball to reach the maximum height is 2 +1 = 3 s.

If u is the velocity of projection and t is the time taken to reach the maximum height, we have (from the equation v = u + at)

0 = u – gt = u – 10×3

This gives u = 30 ms^–1

The maximum height reached is ut – ½ gt² = 30×3 – ½ ×10×3² = 45 m.

(7) A car starts from rest and moves along straight road with uniform acceleration α for some time. Then it moves with uniform retardation β and comes to rest. If the total time of travel is t what is the maximum velocity attained by the car?

(a) 2αβt /(α+ β )

(b) (α+ β )t /αβ

(c) αβt /(α+ β )

(d) αt /(α+ β )

(e) α²βt /(α+ β )

If t₁ is the time interval during which the car was accelerated and v is the maximum velocity of the car during the trip we have the following equations considering the accelerated and decelerated parts of the trip respectively:

v =0 + α t₁ …………….(i)

0 = v – β(t – t₁) ………..(ii)

From these equations, α t₁ = β(t – t₁).

Or, t₁(α+ β ) = βt from which t₁ = βt /(α+ β ).

From Eq.(i) the maximum velocity v = α t₁.

Substituting for t₁, the maximum velocity v = αβt /(α+ β ).

(8) In the above question what is the distance traveled during the decelerated part of the trip?

(a) αβ t² /2(α+ β )²

(b) αβ t² /2(α+ β )

(c) α²β t² /2(α+ β )²

(d) α²β t / (α+ β )²

(e) α²β t² / (α+ β )²

During the decelerated part of the trip the initial velocity is the maximum velocity v which is equal to αβt /(α+ β ). The final velocity is zero. Therefore the distance traveled (s) during the decelerated part of the trip is given by

0 = v² – 2 βs

Thus we have s = v² /2β = [αβt /(α+ β )]² /2β = α²β t² /2(α+ β )²

AP Physics B – Multiple Choice Questions on Kinetic Theory and Thermodynamics

Everything that is really great and inspiring is created by the individual who can labour in freedom

– Albert Einstein

Equations to be remembered in kinetic theory of gases were discussed in the post dated 13^th March 2008, followed by a discussion of some typical multiple choice questions (for practice) in the post dated 17^th march 2008. You can access those posts by clicking on the label ‘kinetic theory’ below this post.

Important points related to thermodynamics were discussed in the post dated 21^st May 2008, followed by a discussion of some typical multiple choice questions (for practice) in the post dated 26^th May 2008. A couple of free response practice questions on thermodynamics were discussed in the post dated 3^rd June 2008. You can access them by clicking on the label ‘thermodynamics’ below this post.

Today we will discuss a few more multiple choice questions on Kinetic Theory and Thermodynamics.

(1) A region of the earth’s atmosphere contains n molecules (treated as ideal gas molecules) per unit volume. The temperature of air in the region is T. If k represents Boltzmann’s constant and R represents universal gas constant, the pressure of air in the region is

(a) nT/k

(b) nkT

(c) RT/n

(d) nRT

(e) nRkT

Since PV = μRT where μ is the number of moles in the volume V of the gas at the pressure P and temperature T, we have

P = (μRT)/V

Since k = R/N where N is Avogadro number, R = Nk.

Therefore, P = (μNkT)/V.

But (μN)/V = n, the number of molecules per unit volume so that P = nkT

[The expression for the pressure P of a gas according to the kinetic theory is

P = (1/3) nmc²

The r.m.s.velocity c of the molecule is √(3kT/m) where m is the molecular mass. On substituting for c in the expression for P you can easily obtain the expression P = nkT].

(2) An ideal monoatomic gas is heated so that it expands at constant pressure. What percentage of the heat supplied to the gas is used to increase the internal energy of the gas?

(a) 100 %

(b) 0 %

(c) 60 %

(d) 50 %

(e) 40 %

Since a monoatomic gas molecule has 3 degrees of freedom, the molar heat capacity (also called molar specific heat) of the monoatomic gas at constant volume (C_V) is (3/2)R where R is universal gas constant. Its molar heat capacity at constant pressure (C_P) is (5/2)R . This follows from Meyer’s relation, C_P = C_V + R.

This means that when heat energy equal to (5/2)R joule is supplied to one mole of a monoatomic gas to make it expand at constant pressure, (3/2)R joule is used to increase the temperature by 1 K and the remaining R joule is used for doing work against the forces which oppose the expansion. 3/2 is 60 % of 5/2. So the correct option is (c).

(3) Oxygen and nitrogen in two enclosures have the same mass, volume and pressure. The ratio of the temperature of oxygen to that of nitrogen is

(a) ⁷/₈

(b) ⁸/₇

(c) 1

(d) ⁴⁹/₆₄

(e) ⁶⁴/₄₉

We have PV = μRT where μ is the number of moles in the volume V of the gas at the pressure P and temperature T.

Since P and V are the same for the two gases,

PV = μ₁RT₁ = μ₂RT₂ where μ₁ and μ₂ are the numbers of moles of oxygen and nitrogen and T₁ and T₂ are their temperatures respectively.

Therefore, μ₁T₁ = μ₂T₂ so that T₁/T₂ = μ₂/μ₁

The samples of the two gases have the same mass (say, m).

Therefore, μ₁ = m/0.032 and μ₂ = m/0.028 since the molar masses of oxygen and nitrogen are 0.032 kg and 0.028 kg respectively.

Therefore, T₁/T₂ = μ₂/μ₁ = 0.032/0.028 = ⁸/₇.

(4) In a Carnot engine 7000 J of heat is absorbed from a source at 400 K and 4900 J of heat is rejected to the sink. The temperature of the sink is

(a) 200 K

(b) 220 K

(c) 260 K

(d) 280 K

(e) 340 K

We have Q₁/T₁ = Q₂/T₂ with usual notations.

Substituting the given values, 7000/400 = 4900/T₂.

Therefore, T₂ = (4900×400)/7000 = 280 K

(5) Three moles of an ideal monoatomic gas is initially in the state A shown in the adjoining pressure-temperature graph. It is taken to state B without changing its pressure. If R is the universal gas constant, the work done by the gas in this process is

(a) 200 R

(b) 300 R

(c) 400 R

(d) 500 R

(e) 600 R

The work done by the gas in taking it from state A to state B = P∆V where ∆V is the increase in volume at constant pressure P.

We have PV = μRT where μ is the number of moles in the sample of the gas and R is the universal gas constant.

Therefore we have

P∆V = μR ∆T = 3×R(450 – 250) = 600R [Option (e)].

Find a few more multiple choice questions (with solution) on thermodynamics here.

AP Physics B&C - Work, Energy & Power - Multiple Choice Questions involving Springs

I am proud of the fact that I never invented weapons to kill

– Thomas Alva Edison

The essential points to be noted in the sections work, energy and power were discussed in the post dated 15^th November 2008. Multiple choice questions in these sections were discussed in the posts dated 20^th November 2008 and 30^th November 2008. Two free response questions were discussed in the post dated 17^th December 2008 You can access all those posts by clicking on the label ‘work’ or ‘energy’ below this post.

Today we will discuss a few multiple choice questions involving springs:

(1) A spring of negligible mass and force constant (spring constant) k contained in a vertical barrel (Fig.) is free to move without friction vertically inside the barrel. A block of mass M, arranged at a height h from the top end of the unstretched spring, falls freely (without friction) from rest on the spring. The maximum compression x of the spring is given by the equation (acceleration due to gravity = g)

(a) mgx = ½ kx²

(b) mgh = ½ kx²

(c) mg (h+x) = ½ k[(h+x)² – x²]

(d) mg (h+x) = ½ k[(h+x)² – h²]

(e) mg (h+x) = ½ kx²

Since the mass M has moved down through a distance h+x when the spring undergoes the maximum compression x, the loss of gravitational potential energy by the mass M is mg (h+x).

The gain in the elastic potential energy by the spring in this condition is ½ kx². The falling mass gains kinetic energy at the cost of gravitational potential energy. When it collides with the spring it transfers its kinetic energy to the spring and compresses the spring. When the spring is compressed to the maximum extent the entire kinetic energy of the mass is converted into elastic potential energy of the spring so that we have

mg (h+x) = ½ kx²

This equation gives the maximum compression x [Option (e)].

In the above question suppose the mass is initially resting on the spring. If the mass is depressed through a distance x and released, the spring will push the mass to a maximum height h (measured from the depressed position) given by the equation,

mgh = ½ kx²

(2) A spring of negligible mass and of force constant (spring constant) 10 Nm^–1 is suspended from the ceiling and it carries a mass M = 0.1 kg (Fig.). A downward force F is applied on this mass so that the spring suffers an additional stretch of 0.1 m. What is the work done by the force F? (Acceleration due to gravity, g = 10 ms^–1).

(a) 0.14 J

(b) 0.15 J

(c) 0.2 J

(d) 0.39 J

(e) 0.1 J

The extension (x₁) of the spring because of the suspended mass = Mg/k = (0.1×10)/10 = 0.1 m.

The work (W) done by the force F is equal to the increase in the total energy of the spring mass system. Note that the elastic potential energy of the spring has increased while the gravitational potential energy of the mass has decreased. Therefore we have

W = ½ k [(x₁ + x₂)² – x₁²] – mgx₂

Or, W = ½ ×10×[0.04 – 0.01] – 0.01 = 0.14 J

(3) The elastic potential energy of a spring stretched by a small distance x is E. The spring is cut into two equal parts and one piece is stretched by the same distance x. The elastic potential energy of the stretched portion is

(a) E

(b) 2E

(c) 4E

(d) E/2

(e) E/4

When the length of a spring is halved, its spring constant is doubled. Its elastic potential energy for a given stretch x is quadrupled since the energy is directly proportional to the square of the stretch.

The correct option is (c).

(4) The adjoining figure shows a block of mass M = 4 kg, moving with uniform velocity along a horizontal frictionless surface AB, towards a spring of force constant k = 100 Nm^–1. One end of the spring is fixed to a rigid vertical wall. The block collides with the spring and produces in it a maximum compression of 0.1 m. What was the momentum of the block before the collision?

(a) 0.5 kg ms^–1

(b) 2 kg ms^–1

(c) 2.5 kg ms^–1

(d) 4 kg ms^–1

(e) 8 kg ms^–1

Since the spring is compressed by 0.1 m, the elastic potential energy (which is ½ kx² where x is the compression) of the spring is ½ ×100×(0.1)² = 0.5 J.

Since the the block imparted this energy to the spring, the kinetic energy of the block was 0.5 J.

If p is the momentum of the block, we have

p²/2M = 0.5

Or, p²/(2×4) = 0.5 from which p = 2 kg ms^–1

You can find similar useful questions (with solution) in this section here.